JEE Main 2016 Question Paper With Solutions (3rd-April-2016)


90 Questions MCQ Test JEE Main Mock Test Series 2020 & Previous Year Papers | JEE Main 2016 Question Paper With Solutions (3rd-April-2016)


Description
This mock test of JEE Main 2016 Question Paper With Solutions (3rd-April-2016) for JEE helps you for every JEE entrance exam. This contains 90 Multiple Choice Questions for JEE JEE Main 2016 Question Paper With Solutions (3rd-April-2016) (mcq) to study with solutions a complete question bank. The solved questions answers in this JEE Main 2016 Question Paper With Solutions (3rd-April-2016) quiz give you a good mix of easy questions and tough questions. JEE students definitely take this JEE Main 2016 Question Paper With Solutions (3rd-April-2016) exercise for a better result in the exam. You can find other JEE Main 2016 Question Paper With Solutions (3rd-April-2016) extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be :

Solution:


Mean deviation = 
L. C. = 1 s.
∴ Required value = 92 ± 2 s

*Multiple options can be correct
QUESTION: 2

A particle of mass m is moving along the side of a square of side ‘a’, with a uniform speed v in the x-y plane as shown in the figure :

Which of the following statements is false for the angular momentum L about the origin? 

Solution:

Along CD, ⊥ distance from line of motion =
∴ Magnitude of angular momentum =
Hence (2) is incorrect. 
In option (4) the direction of is incorrect. 

QUESTION: 3

A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals m. The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR.
The values of the coefficient of friction m and the distance x (= QR), are, respectively close to : 

Solution:

From work energy theorem and given condition

∴ 

∴ 

QUESTION: 4

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8 x 107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms-2 :

Solution:

0.2 x 3.8 x 107 x m = 10 x g x 1 x 1000
m =  = 1.289 x 10-2 Kg
= 12.89 x 10-3 Kg

QUESTION: 5

A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD which are placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and CD (see figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to:

Solution:

If we take ‘r’ as the distance of IAOR from the axis of rotation, then ‘r’ decreases on left side as the object moves forward.
So, for left v = ωr' < ωr (for right point)
So, the roller will turn to the left as it moves forward.

QUESTION: 6

A satellite is revolving in a circular orbit at a height ‘h’ from the earth’s surface (radius of earth R; h <<R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to : (Neglect the effect of atmosphere).

Solution:

Orbital velocity v = 
Velocity required to escape

∴ Increase in velocity

QUESTION: 7

A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4s a day if the temperature is 20°C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion (α) of the metal of the pendulum shaft are respectively:

Solution:

Time period of pendulum,



Thus, the coefficient of linear expansion in pendulum clock = 1.85 x 10-5/C

QUESTION: 8

An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity Cremains constant. If during this process the relation of pressure P and volume V is given by PVn = constant, then n is given by (Here CP and CV are molar specific heat at constant pressure and constant volume, respectively) :

Solution:

QUESTION: 9

‘n’ moles of an ideal gas undergoes a process A → B as shown in the figure. The maximum temperature of the gas during the process will be :

Solution:

Equation of line AB



(For maximum temperature) 

(Condition for maximum temperature)

QUESTION: 10

A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance 2A/3 from equilibrium position. The new amplitude of the motion is :

Solution:

2 = k
Total initial energy = 1/2 kA2
at x = 2A/3, potential energy = 
Kinetic energy at

If speed is tripled, new Kinetic energy = 
∴ New total energy = 
If next amplitude = A';  then 

QUESTION: 11

A uniform string of length 20 m is suspended from a rigid support. A short wave pulse in introduced at its lowest end. It starts moving up the string. The time taken to reach the support is:
(take g = 10 ms-2)

Solution:


QUESTION: 12

The region between two concentric spheres of radii ‘a’ and ‘b’, respectively (see figure), has volume charge density ρ = A/r, where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is :

Solution:

Charge in the shaded region = 
Total field at P = 

For field to be independent of r : Q = 2πAa2

QUESTION: 13

A combination of capacitors is set up as shown in the figure. The magnitude of the electric field due to a point charge Q (having a charge equal to the sum of the charges on the 4 μF and 9 μF capacitors), at a point distant 30 m from it, would equal:

Solution:


Charges on 3 μF = 3 μF x δV = 24 μC 
∴ Charge on 4 μF = Charge on 12 μF = 24 μC
Charge of 3 μF = 3 μF ´ 2V = 6 μC
Charge of 9 μF = 9 μF ´ 2V = 18 mC
Charge on 4 μF + Charge on 9 μF = (24 + 18) μF = 42 μC
∴ Electric field at 30 m = 9 x 103 

QUESTION: 14

The temperature dependence of resistances of Cu and undoped Si in the temperature range 300-400 K, is best described by :

Solution:

QUESTION: 15

Two identical wires A and B, each of length  ‘’, carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side ‘a’. If BA and BB are the values of magnetic field at the centres of the circle and square respectively, then the ratio BA / Bis:

Solution:




QUESTION: 16

Hysteresis loops for two magnetic materials A and B are given below :
These materials are used to make magnets for electric generators, transformer core and electromagnet core.
Then it is proper to use :

Solution:

Conceptual (Requires low retentivity and low coercivity)

QUESTION: 17

An arc lamp requires a direct current of 10 A at 80 V to function. It is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to:

Solution:





∴ 

QUESTION: 18

Arrange the following electromagnetic radiations per quantum in the order of increasing energy :
A : Blue light  
B : Yellow light  
C : X-ray  
D : Radiowave

Solution:

QUESTION: 19

An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observer the tree appears :

Solution:

QUESTION: 20

The box of a pin hole camera, of length L, has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength λ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say bmin) when :

Solution:





[By substituting for a from (ii) in (i)] 
∴ The radius of the spot = 

QUESTION: 21

Radiation of wavelength λ, is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is changed to 3λ/4, the speed of the fastest emitted electron will be:

Solution:

.... (i)
 ...... (ii)
(ii) - (i) gives


.... (iv)
combining (iii) & (iv)

QUESTION: 22

Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the samples have equal number of nuclei. After 80 minutes, the ratio of decayed numbers of A and B nuclei will be :

Solution:

80 minutes = 4 half-lives of A = 2 half-lives of B
Let the initial number of nuclei in each sample be N
NA after 80 minutes = N/2⇒ Number of A nuclides decayed = 15/16N
NB after 80 minutes = N/2⇒ Number of B nuclides decayed = 3/4N
Required ratio = 

QUESTION: 23

If a, b, c, d are inputs to a gate and x is its output, then, as per the following time graph, the gate is :

Solution:

x is 1 when at least one of the inputs is 1. Hence x is an OR-Gate.

QUESTION: 24

A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge and brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?

Solution:

L.C. = 
-ve zero error = - 5 x L.C. = -0.005 mm
∴ Measured value
= main scale reading + screw gauge reading - zero error 
= 0.5 mm + {25 x 0.001 - (-0.05)} mm  = 0.80 mm

QUESTION: 25

Choose the correct statement :

Solution:

QUESTION: 26

A pipe open at both ends has a fundamental frequency f in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now :

Solution:

QUESTION: 27

A galvanometer having a coil resistance of 100 Ω gives a full scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10 A is :

Solution:

Maximum voltage that can be applied across the galvanometer coil = 100 Ω x 10-3 A = 0.1 V.
If RS is the shunt resistance :
RS x 10 A = 0.1 V
⇒ RS = 0.01 Ω

QUESTION: 28

In an experiment for determination of refractive index of glass of a prism by i - δ, plot, it was found that a ray incident at angle 35°, suffers a deviation of 40° and that it emerges at angle 79°. In that case which of the following is closest to the maximum possible value of the refractive index?

Solution:


δ =  i + e - A
40 = 35 + 79 - A
40 = 114 - A
A = 114 - 40 = 74 = r1+ r2 
From this we get,
μ = 1.5
∴ δmin < 40°

QUESTION: 29

Identify the semiconductor devices whose characteristics are given below, in the order (a), (b), (c), (d):

Solution:

Graph (a) is for a simple diode.
Graph(b) is showing the V Break down used for zener diode.
Graph (c) is for solar cell which shows cut-off voltage and open circuit current
Graph (d) shows the variation of resistance h and hence current with intensity of light.

QUESTION: 30

For a common emitter configuration, if a and b have their usual meanings, the incorrect relationship between a and b is:

Solution:

QUESTION: 31

At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is:

Solution:




QUESTION: 32

Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure pi and temperature T1 are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to T2. The final pressure pf is :

Solution:

Initially

n1 = piV/ RT1
n2 = piV/ RT1
Later on

n'1 = pfV/ RT1
n'2 = pfV/ RT2

QUESTION: 33

A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h/l (where l is wavelength associated with electron wave) is given by :

Solution:



=> 

QUESTION: 34

The species in which the N atom is in a state of sp hybridization is :

Solution:

0 ==0
sp hybridisation 

QUESTION: 35

The heats of combustion of carbon and carbon monoxide are - 393.5 and -283.5 kJ mol-1, respectively.
The heat of formation (in kJ) of carbon monoxide per mole is :

Solution:

  

QUESTION: 36

18 g glucose (C6H12O6) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is :

Solution:

QUESTION: 37

The equilibrium constant at 298 K for a reaction A + BC + D is 100. If the initial concentration of all the four species were 1M each, then equilibrium concentration of D (in mol L-1) will be : 

Solution:


QUESTION: 38

Galvanization is applying a coating of :

Solution:

Galvanization is applying a coating of zinc.

QUESTION: 39

Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be :

Solution:


Rate of H2O2 decomposition = 

QUESTION: 40

For a linear plot of log (x/m) versus log p in a Freundlich adsorption isotherm, which of the following statements is correct ? (k and n are constants)

Solution:


QUESTION: 41

Which of the following atoms has the highest first ionization energy ?

Solution:


Scandium has the highest first Ionisation energy.

QUESTION: 42

Which one of the following ores is best concentrated by froth floatation method ?

Solution:

PbS, i.e. Galena is best concentrated by froth floatation method.

QUESTION: 43

Which one of the following statements about water is FALSE ?

Solution:

Water possess intermolecular hydrogen bonding in the condensed phase.

QUESTION: 44

The main oxides formed on combustion of Li, Na and K in excess of air are, respectively :

Solution:

Li --> Li2O
Na --> Na2O2
K --> KO2

QUESTION: 45

The reaction of zinc with dilute and concentrated nitric acid, respectively produces :

Solution:

Zn reacts with dil. HNO3 (20%) to form nitrous oxide (N2O)
Zn reacts with conc. HNO3 (70%) to form nitrogen dioxide (NO2)
4Zn + 10HNO3(dil.) --> 4Zn(NO3)2 + N2O + 5H2O
Zn + 4HNO3 (conc.) --> Zn(NO3)2 + 2NO2 + 2H2O

QUESTION: 46

The pair in which phosphorous atoms have a formal oxidation state of + 3 is :

Solution:

Phosphorus acid series contain phosphorus in the oxidation state (+ III).

        Pyrophosphorus acid


      orthophosphorus acid

QUESTION: 47

Which of the following compounds is metallic and ferromagnetic ?

Solution:

CrO2 is metallic and ferromagnetic in nature.

QUESTION: 48

The pair having the same magnetic moment is :
[At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27]

Solution:

Cr+2 -->(Ar) 3d4 4s

 

Fe+2 --> (Ar) 3d64s0

QUESTION: 49

Which one of the following complexes shows optical isomerism?

Solution:

QUESTION: 50

The concentration of fluoride, lead, nitrate and iron in a water sample from an underground lake was found to be 1000 ppb, 40 ppb, 100 ppm and 0.2 ppm, respectively. This water is unsuitable for drinking due to high concentration of :

Solution:

Fluoride - 1000 ppm
Lead - 40 ppm
Nitrate - 100 ppm
[> 40 ppm gives blue baby syndrome disease.]
Iron - 0.2 ppm

QUESTION: 51

The distillation technique most suited for separating glycerol from spent-lye in the soap industry is :

Solution:

Glycerol and spent-lye can be separated by distillation under reduced pressure.

QUESTION: 52

The product of the reaction given below is :

Solution:

QUESTION: 53

The absolute configuration of
 is : 

Solution:

QUESTION: 54

2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields :
                          

Solution:


QUESTION: 55

The reaction of propene with HOCl (Cl2 + H2O) proceeds through the intermediate:

Solution:

QUESTION: 56

In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole of amine produced are :

Solution:

QUESTION: 57

Which of the following statements about low density polythene is FALSE ?

Solution:

QUESTION: 58

Thiol group is present in :

Solution:

QUESTION: 59

Which of the following is an anionic detergent ?

Solution:

CH3(CH2)11OSO3Na  = (Sodium Lauryl Sulphate)

QUESTION: 60

​The hottest region of Bunsen flame shown in the figure below is :

Solution:

Region - 2 --> blue flame

QUESTION: 61

If f(x) + 2f(1/x) = 3x, x ≠ 0, and S = {x ∈ R : f(x) = f(-x)}; then S: 

Solution:


QUESTION: 62

A value of θ for which  is purely imaginary, is: 

Solution:


To be purely imaginary if

QUESTION: 63

The sum of all real values of x satisfying the equation = 1 is:

Solution:

& base x2 - 5x + 5 = 0  or  1 or -1
If  x2 - 5x + 5 = 0
But it will not satisfy original equation.
 
x2 - 5x + 5 = -1
∴ x = 2, 3
x = 3 does not satisfy eqn.
Hence solutions are -10, 6, 4, 1, 2
So, sum of solutions = -10 + 6 + 4 + 1 + 2 = 3 

QUESTION: 64

If A = and A adj A = A AT, then 5a + b is equal to:

Solution:

Equate,  10a + 3b = 25a2 + b2
& 10a + 3b = 13
& 15a - 2b = 0
a/2 = b/15 = k (let) 
Solving a = 2/5, b = 3 
So, 5a + b = 5 x 2/5 + 3 = 5

QUESTION: 65

The system of linear equations
x + λy - z = 0
λx - y - z = 0
x + y - λz = 0
​has a non-trivial solution for:

Solution:

x + λy - z = 0
λx - y - z = 0
x + y - λz = 0
For non-trivial solution = > Δ = 0
 = 0

Exactly 3 values of λ. 

QUESTION: 66

If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is:

Solution:

  A    LL MS 

Total words = 12 + 24 + 12 + 3 + 6 = 57 SMALL   58th
∴ the position of the word SMALL is 58th.

QUESTION: 67

​If the sum of the first ten terms of the series
m, then m is equal to: 

Solution:

upto 10 terms
upto 10 terms.
(8)2 + (12)2 + (16)2 + ……… up to 10 terms
Tn = [4 (n + 1)]2 where n varies from 1 to 10.
= 16(n2 + 2n + 1)

∴ upto 10 terms =
It is given that = 16/5 m
∴ m = 505/5 = 101

QUESTION: 68

If the number of terms in the expansion of , x ≠ 0, is 28, then the sum of the coefficients of all the terms in this expansion, is: 

Solution:

Number of terms = = 28
=> n = 6

Put x = 1, n = 6, a0 + a1 + a2 + …+a2n = 36 = 729

QUESTION: 69

Let p = then loge p is equal to: 

Solution:

QUESTION: 70

If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is:

Solution:

t2 = a + d
t5 = a + 4d
t9 = a + 8d
Given  t2, t5, t9 are in G.P.
(a + 4d)2 = (a + d) (a + 8d)
a2 + 16d2 + 8ad = a2 + 8d2 + 9ad
8d2 - ad = 0
d(8d - a) = 0
As given non-constant AP. => d ≠ 0
∴ d = a/8 => a = 8d
so,  A.P. is  8d, 9d, 10d, …..
Common ratio of G.P. = 

QUESTION: 71

​For x ∈ R, f(x) = |log 2 - sin x| and g(x) = f(f(x)), then:

Solution:

QUESTION: 72

Consider  
A normal to y = f(x) at x = π/6 also passes through the point: 

Solution:


QUESTION: 73

A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then:

Solution:

Let length of two parts be ‘a’ and ‘2 - a’ As per condition given, we write
a = 4x and 2 - a = 2πr




 

QUESTION: 74

The integral dx is equal to:

Solution:


Dividing numerator and denominator by x15 we get,

QUESTION: 75

is equal to: 

Solution:



P = 27/e2

QUESTION: 76

The area (in sq. units) of the region {(x, y) : y2 ≥ 2x and x2 + y2 ≤ 4x, x ≥ 0, y ≥ 0} is:

Solution:


x2 + y2 ≤ 4x  &  y2 ≥ 2x
To find point of intersection,
x2 + y2 = 4x => x2 + 2x = 4x
=> x2 - 2x = 0 => x (x - 2) = 0
=> x = 0  or  x = 2
∴ y = 0 or y = 2
Solve (x, y) = (0, 0) & (x, y) = (2, 2)
Area = 

QUESTION: 77

If a curve y = f(x) passes through the point (1, -1) and satisfies the differential equation, y(1 + xy) dx = x dy, then is equal to: 

Solution:



We have to find 
Put x = 

QUESTION: 78

Two sides of a rhombus are along the lines, x - y + 1 = 0 and 7x - y - 5 = 0. If its diagonals intersect at (-1, -2), then which one of the following is a vertex of this rhombus?

Solution:

Coordinates of A = (1, 2)
∴ Slope of AE = 2
=> Slope of BD =
=> Eq. of BD is 
=> x + 2y + 5 = 0
Co-ordinates of D = 

QUESTION: 79

The centres of those circles which touch the circle, x2 + y2 - 8x - 8y - 4 = 0, externally and also touch the x-axis, lie on:

Solution:

x2 + y2 - 8x - 8y - 4 = 0
Centre (4, 4)
Radius =  = 6
Let centre of the circle is (h, k)
= (6 + k)
(h - 4)2 + (k - 4)2 = (6 + k)2
h2 - 8h + 16 + k2 - 8k + 16 = 36 + k2 + 12k
h2 - 8h - 20k - 4 = 0
x2 - 8x - 20y - 4 = 0
Which is an equation of parabola

QUESTION: 80

If one of the diameters of the circle, given by the equation, x2 + y2 - 4x + 6y - 12 = 0, is a chord of a circle S, whose centre is at (-3, 2), then the radius of S is:

Solution:

x2 + y2 - 4x + 6y - 12 = 0
Centre (2, -3)
Radius = 5
Distance b/w two centres c1(2, -3) and c2(-3, 2)
d = 
Radius of (S) = 

QUESTION: 81

Let P be the point on the parabola, y2 = 8x which is at a minimum distance from the centre C of the circle, x2 + (y + 6)2 = 1. Then the equation of the circle, passing through C and having its centre at P is:

Solution:

Normal at P(at2, 2at) is y + tx = 2at + at3 Given it passes (0, -6)
=> -6 = 2at + at(a = 2)
-6 = 4t + 2t3
t3 + 2t + 3 = 0
t = -1
so, P (a, -2a) = (2, -4) . [a = 1)
radius of circle = CP = 
Circle is (x - 2)2 + (y + 4)2
x2 + y2 - 4x + 8y + 12 = 0

QUESTION: 82

The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half the distance between its foci, is:

Solution:


Squaring eqn. (2), we get 
and we know that 

QUESTION: 83

The distance of the point (1, -5, 9) from the plane x - y + z = 5 measured along the line x = y = z is :

Solution:

Let Q(1, -5, 9)

Line is (say)
Any pt on line we can take P(r + 1, r - 5, r + 9)
So, Pt satisfy Plane
=>(r + 1) - (r - 5) + (r + 9) = 5 r = -10 So, Point P = (-9, -15, -1)
Distance is PQ = 

QUESTION: 84

If the line,  lies in the plane, lx + my - z = 9, then l2 + m2 is equal to : 

Solution:

Point on line is P = (3, -2, -4)
'P’ lies on lx + my - z = 9
=> 3l - 2m + 4  = 9
3l - 2m = 5 ....(1)
As line lies on plane
=> 2 x l + m x (-1) + 3 x (-1) = 0
2l - m = 3 .....(3) 
Solving l = 1, m = -1
So, l+ m= 2

QUESTION: 85

Let andbe three unit vectors such that If is not parallel to c , then the angle betweenandis :

Solution:


Equate,

 as unit vectors

 

QUESTION: 86

If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true?

Solution:


=> (4a2 + 536) - (a2 + 32a + 256) = 196 => 3a2 - 32a + 84 = 0

QUESTION: 87

Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up four, E2 is the event that die B shows up two and E3 is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true?

Solution:

P(E1) = 1/6
P(E2) = 1/6
P(E3) = 1/2
P(E1E2) = 1/36
P(E2E3) = 1/12
P(E3E1) = 1/12
P(E1 E2 E3) = 0
∴ E1, E2, E3 are not independent

QUESTION: 88

If 0 ≤ x < 2π, then the number of real values of x, which satisfy the equation cos x + cos 2x = cos 3x + cos 4x = 0, is

Solution:

We have, cosx + cos2x + cos 3x + cos 4x = 0
(cos x + cos 4x)+ (cos 2x+ cos 3x)= 0



Or


Solution are 


…  (0 ≤ x < 2p) 

QUESTION: 89

A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30°. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60°. Then the time taken (in minutes) by him, from B to reach the pillar, is :

Solution:


Let AB = x, BQ = y, PQ = z

x = 2y
∴ y =x/2
To go x, it takes 10 minutes.
∴ To go y, it takes 5 minutes.

QUESTION: 90

The Boolean Expressionis equivalent to :

Solution:

(p ∧ ~ q) ν q ν (~ p ∧ q)